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Suppose $X_1, ..., X_n$ are i.i.d. random variables uniformly distributed in the unit ball in $\mathbb{R}^m$. What is the expected number of vertices that their convex hull has?

The only thing I managed to prove here was:

$$P(\text{ convex hull of }X_1, ..., X_n\text{ has exactly }k\text{ vertices}) = C_n^k P(X_{k+1}, ... ,X_n \text{ lie in the convex hull of }X_1, ..., X_n)$$

Not sure, however, whether this helps or not.

Chain Markov
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  • From the displayed equation, it seems that you wanted the last variable in the first line to be $X_n$, not $X_m$? – joriki Dec 09 '19 at 09:18
  • According to this, the expected number is $O\left(n^{\frac{m-1}{m+1}}\right)$. See also https://math.stackexchange.com/questions/346090 and https://mathoverflow.net/questions/37578. – joriki Dec 09 '19 at 09:21

1 Answers1

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The expected number of vertices in the convex hull is $O(n^{\frac{m - 1}{m + 1}})$ for $n \to \infty$. It was proved in 1970 by H. Raynaud in "Sur l’enveloppe convex des nuages de points aleatoires dans $\mathbb{R}_n$".

Chain Markov
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  • The exponent approaches 1 as the spatial dimension $m$ increases, so at a very high dimension, all points are at the surface (i.e., on the convex hull). Isn't this counter-intuitive? – Taozi Oct 14 '21 at 20:49
  • @Taozi, yes, when the number of dimensions become too high, weird things tend to happen. Richard Bellman called such effects “the curse of dimensionality”. – Chain Markov Oct 25 '21 at 21:11
  • Thanks for the keywords, I had a good time reading. – Taozi Oct 26 '21 at 01:40